Question

find the equation of the straight line perpendicular to the two lines x+1 /-3 = y-3 /2 = z+2 /1 and x / 1 =y -7 /- 3 = z+7 /2 and passing through the point of intersection.

Answer 1

Therefore the equation of straight line is $\frac{x-2}{1} =\frac{y-1}{1} =\frac{z+3}{1}$

Step-by-step explanation:

Given equations of straight line are

$\frac{x+1}{-3} =\frac{y-3}{2} =\frac{z+2}{1}$............(1)

and $\frac{x}{1} =\frac{y-7}{-3} =\frac{z+7}{2}$..........(2)

Let the direction of ratio of the straight line which is perpendicular to given straight line be l,m ,n

Since the required line is perpendicular to the given lines so

-3l+2m+n=0..........(3)

l-3m+2n=0.............(4)

From above two equation we get

$\frac{l}{7} =\frac{m}{7} =\frac{n}{7}=k(say)$

Therefore the direction ratio of the required line is (7k,7k,7k) =(1,1,1)

Any point of equation (1) be P(-3r-1,2r+3,r-2)

Any point of equation (2) be Q (s, -3s+7,2s-7)

Let the intersection point of the given lines be P and Q

So, - 3r - 1 =s ,  2r+3 = -3s +7  and r-2 = 2s-7

Its gives r= - 1 and s = 2

So the coordinate of intersection point of given two lines is=(2,1,-3)

Therefore the equation of straight line which passes through the point (2,1,-3) and with direction ratio (1,1,1) is

$\frac{x-2}{1} =\frac{y-1}{1} =\frac{z+3}{1}$